Suppose the functions \( f, g, h \) and \( l \) are defined as follows: \[ \begin{aligned} f(x) & =4 x^{2}-5 x+1 \\ g(x) & =2 \sqrt{2-\frac{x}{2}}-x \\ h(x) & =-\frac{1}{2} x+3 \\ l(x) & =\log _{4}(x+3)-\log _{4}(x-2)\end{aligned} \] \[ \begin{array}{l}\text { Solve the equation } g(x)=-4\end{array} \]

To solve the equation \( g(x) = -4 \) where \( g(x) = 2 \sqrt{2 - \frac{x}{2}} - x \), we start by substituting \(-4\) into the function: \[ 2 \sqrt{2 - \frac{x}{2}} - x = -4 \] Now, isolate the square root: \[ 2 \sqrt{2 - \frac{x}{2}} = -4 + x \] To simplify, divide both sides by 2: \[ \sqrt{2 - \frac{x}{2}} = \frac{-4 + x}{2} \] Next, square both sides to eliminate the square root (keeping in mind that this can potentially introduce extraneous solutions): \[ 2 - \frac{x}{2} = \left(\frac{-4 + x}{2}\right)^2 \] Now, expand the right side: \[ 2 - \frac{x}{2} = \frac{(-4 + x)^2}{4} \] Multiply through by 4 to clear the fraction: \[ 8 - 2x = (-4 + x)^2 \] Expanding the square gives: \[ 8 - 2x = 16 - 8x + x^2 \] Rearranging all terms to one side: \[ 0 = x^2 - 6x + 8 \] Now we can factor the quadratic: \[ 0 = (x - 2)(x - 4) \] Setting each factor to zero gives the possible solutions: \[ x - 2 = 0 \quad \Rightarrow \quad x = 2 \] \[ x - 4 = 0 \quad \Rightarrow \quad x = 4 \] Next, we need to check both potential solutions in the original function \( g(x) \) to ensure they do not introduce any extraneous solutions. **For \( x = 2 \):** \[ g(2) = 2 \sqrt{2 - \frac{2}{2}} - 2 = 2 \sqrt{2 - 1} - 2 = 2 \cdot 1 - 2 = 0 \quad (\text{not } -4) \] **For \( x = 4 \):** \[ g(4) = 2 \sqrt{2 - \frac{4}{2}} - 4 = 2 \sqrt{2 - 2} - 4 = 2 \cdot 0 - 4 = -4 \quad (\text{solution!}) \] Thus, the only solution to the equation \( g(x) = -4 \) is: \[ \boxed{4} \]